Simple Harmonic Motion - Velocity EQ.

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Homework Help Overview

The discussion revolves around the velocity equation for an object undergoing simple harmonic motion, specifically given the position function x = 0.20m*sin(0.44 rad/s * t). Participants are exploring how to derive the velocity from this position function and questioning the validity of the provided solutions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the correct form of the velocity equation, with some asserting that the derivative of the position function should yield a cosine function, while others question the phase shift in the equation. There are inquiries about the source of the velocity equation presented in the textbook and the implications of using radians versus degrees in calculations.

Discussion Status

The discussion is ongoing, with multiple interpretations of the velocity equation being explored. Some participants have provided guidance on deriving the velocity from the position function, while others express confusion regarding the phase shift and its impact on the solution. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can share. The original poster's attempt at deriving the velocity equation has led to confusion, particularly regarding the phase shift and the use of radians versus degrees in calculations.

Masri
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Homework Statement



x = 0.20m*sin(0.44 rad/s * t)


Write the equation for the velocity of this object. Find the velocity at t = 4:0 s.

Homework Equations





The Attempt at a Solution



well, i know that v(t)= - wAsin(wt +\phi)

and we know the w = 0.44 rad/s
A = 0.20 s
t=4

so it i plugged the numbers in but the answer came wrong..

and the answer in the book says ...
v= 0.088 m/s* cos ( 0.44 rad/s * 4 s)

v= - 1.7 *10-2 m/s


well first where did he get that equation from ? 2nd if we actually calculated (v= 0.088 m/s* cos ( 0.44 rad/s * 4 s) ) it won't give us v= - 1.7 *10-2 m/s

so if someone could please help me out with this one... any help would be appreciated ,

Thanks!
 
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Masri said:
and the answer in the book says ...
v= 0.088 m/s* cos ( 0.44 rad/s * 4 s)

v= - 1.7 *10-2 m/s


As i know, the calculation of the book was using radian(0.44) changed to degree(25.120).So,the answer would get -0.01655 m/s.

But for me, normally we usage radian.:smile:
 
but where did he get that formula from ?
 
hello,

your equation for the velocity is incorrect. can you show what you did to derive it?
 
Masri said:
but where did he get that formula from ?
(0.44rad/pi)x1800=25.120
 
eczeno said:
hello,

your equation for the velocity is incorrect. can you show what you did to derive it?

it is actually given to us like this in our formula sheet . (am talking about the on i used, not the one the book )


thanks xiaoB, but i was talking about the velocity equation that the book used
 
eczeno said:
hello,

your equation for the velocity is incorrect. can you show what you did to derive it?

Hi,eczeno!:smile:

Actually he correct because the sine has the phase ∅.
 
if the formula you gave for the position is correct, then your velocity is incorrect.
v = dx/dt. take the derivative of x to find v.

notice the solution has cosine where you have sine; the solution given is correct.
 
Masri said:
it is actually given to us like this in our formula sheet . (am talking about the on i used, not the one the book )


thanks xiaoB, but i was talking about the velocity equation that the book used

You just derive this x = 0.20m*sin(0.44 rad/s * t) respected for t :

dx/dt=V.
 
  • #10
xiaoB said:
Hi,eczeno!:smile:

Actually he correct because the sine has the phase ∅.

No, it's not correct. The derivative of A*sin(wt) is wA*cos(wt).
 
  • #11
Dick said:
No, it's not correct. The derivative of A*sin(wt) is wA*cos(wt).

Yes ! the derive should be this wA*cos(wt) .

But if v(t)= - wAsin(wt + ∅) the sin(wt + ∅) for the phase ∅ is 900 so the sin(wt + 90)=cos(wt).:smile:

Therefore,v(t)= - wAsin(wt + ∅) nothings going wrong.:smile:
 
  • #12
xiaoB said:
Yes ! the derive should be this wA*cos(wt) .

But if v(t)= - wAsin(wt + ∅) the sin(wt + ∅) for the phase ∅ is 900 so the sin(wt + 90)=cos(wt).:smile:

Therefore,v(t)= - wAsin(wt + ∅) nothings going wrong.:smile:

of course the addition of the proper phase will turn cos into sin, but there is no phase specified in the attempted solution, so it is incorrect.

this is really only adding unnecessary confusion to a relatively simple problem. the derivative of sin is cos; this gets us directly to the correct solution without needing to worry about the phase.
 
  • #13
xiaoB said:
Yes ! the derive should be this wA*cos(wt) .

But if v(t)= - wAsin(wt + ∅) the sin(wt + ∅) for the phase ∅ is 900 so the sin(wt + 90)=cos(wt).:smile:

Therefore,v(t)= - wAsin(wt + ∅) nothings going wrong.:smile:

If I were first learning simple harmonic motion, that sort of approach to solving the problem would make me quit physics. That's what's going wrong.
 
  • #14
Dick said:
If I were first learning simple harmonic motion, that sort of approach to solving the problem would make me quit physics. That's what's going wrong.

Oh!thanks.I not think so...
 

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